Intermediate Algebra (11th edition)

(Marvins-Underground-K-12) #1

OBJECTIVE 4 Solve an equation that is quadratic in form by substitution.


A nonquadratic equation that can be written in the form


for and an algebraic expression u,is called quadratic in form.


Many equations that are quadratic in form can be solved more easily by defining


and substituting a “temporary” variable ufor an expression involving the variable in


the original equation.


Defining Substitution Variables

Define a variable u, and write each equation in the form


(a)


Look at the two terms involving the variable x, ignoring their coefficients. Try to


find one variable expression that is the square of the other. Since we can


define and rewrite the original equation as a quadratic equation.


Here,

(b)


Because this equation involves both and we choose


Substituting ufor gives the quadratic equation


. Here,


(c)


We apply a power rule for exponents (Section 5.1), Because


we define The original equation becomes


Here, NOW TRY

Solving Equations That Are Quadratic in Form

Solve each equation.


(a)


We can write this equation in quadratic form by substituting (See


Example 5(a).)


Let
Factor.

or Zero-factor property


or Solve.


or Substitute for u.


or Square root property


The equation , a fourth-degree equation, has four solutions,


* The solution set is abbreviated Each solution can be veri-


fied by substituting it into the original equation for x.


- 3, -2, 2, 3. 5 2,  36.


x^4 - 13 x^2 + 36 = 0


x=  2 x= 3


x^2 = 4 x^2 = 9 x^2


u= 4 u= 9


u- 4 = 0 u- 9 = 0


1 u- 421 u- 92 = 0


u^2 - 13 u+ 36 = 0 u=x^2.


1 x^222 - 13 x^2 + 36 = 0 x^4 = 1 x^222


x^4 - 13 x^2 + 36 = 0


u for x^2.


x^4 - 13 x^2 + 36 = 0


EXAMPLE 6


2 u^2 - 11 u+ 12 = 0. u=x1/3.


1 x1/3 22 = x2/3, u=x1/3.


1 am 2 n= amn.


2 x2/3- 11 x1/3 + 12 = 0


2 u^2 + 7 u+ 5 = 0 u= 4 x-3.


u= 4 x- 3. 4 x- 3


14 x- 322 14 x- 32 ,


214 x- 322 + 714 x- 32 + 5 = 0


u^2 - 13 u+ 36 = 0 u=x^2.


u= x^2 ,


x^4 = 1 x^222 ,


x^4 - 13 x^2 + 36 = 0


au^2 +bu+c= 0.


EXAMPLE 5


aZ 0


au^2 buc0,


516 CHAPTER 9 Quadratic Equations, Inequalities, and Functions


*In general, an equation in which an nth-degree polynomial equals 0 has ncomplex solutions, although some
of them may be repeated.

Don’t stop here.

NOW TRY
EXERCISE 5
Define a variable u, and write
each equation in the form


(a)


(b)



  • 111 x+ 22 + 4 = 0


61 x+ 222

x^4 - 10 x^2 + 9 = 0

au^2 +bu+c=0.


NOW TRY ANSWERS



  1. (a)
    (b)
    6 u^2 - 11 u+ 4 = 0


u=x+2;

u=x^2 ; u^2 - 10 u+ 9 = 0
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